Different approaches when working with the product topology on $\mathbb{R}^{\mathbb{N}}$

Question

Consider the space $\Bbb R^\Bbb N$ with the product topology. If $B$ is the set of bounded sequences in $\Bbb R^\Bbb N$ show that $B$ is neither open or closed.

I have few ideas for this and I would like to validate whether they both accomplish the same thing. So the question is more related to the product topology than the actual problem.

The main idea seems to be here that the basic open sets in the product topology only restrict finitely many indices and after some index all other values will become arbitary?

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To show that $B$ is not open in the product topology I would need to show that for any sequence of functions $(f_n)_{n \in \Bbb N}$ there isn't a neighborhood $U$ such that for all $(g_n)_{n \in \Bbb N} \in U$ we would have that $(g_n)_{n \in \Bbb N} \in B$ right?

Now as we can restrict ourselves to basic open sets we would have to satisfy that if $$(f_n)_{n \in \Bbb N} \in \prod_{n \in \Bbb N} U_n$$ then for $(g_n)_{n \in \Bbb N} \in \prod_{n \in \Bbb N} U_n$ it would need to be that $(g_n)_{n \in \Bbb N} \in B$ that is $(g_n)_{n \in \Bbb N}$ is bounded. However with the product topology we have that $U_n \ne \Bbb R$ for only finitely many indices that is $$\prod_{n \in \Bbb N} U_n = U_1 \times U_2 \times \dots \times \Bbb R \times \Bbb R \times \dots$$ so the sequence $(g_n)_{n \in \Bbb N}$ has only finitely many components that are bounded. Am I corret to claim that this is enough to show that $(g_n)_{n \in \Bbb N}$ is not neccessarliy bounded?

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The second approach is that it would be satisfactory to show that I can find a bounded sequence of functions $(f_n)_{n \in \Bbb N}$ that converges to non-bounded sequence of functions.

First off to say that a sequence of functions is bounded I think it means that for any $n \in \Bbb N$ and for any $k \in \Bbb N$ we have that $|f_n(k)| < M$ for some $M$ always?

If so then the sequence $(f_1, f_2, \dots)$ where $f_1=(1,0,0, \dots), f_2=(1,2, 0, 0, \dots)$ etc converges to $(1,2,3,4,5, \dots)$ that is not bounded, but each $f_i$ is. This shows then that there is a sequence of bounded functions that converge non-bounded sequence.

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These two approaches are possibly very much alike, but I don't see how the latter one uses the product topology in any way? Are these two equivalent or is there some major differences?

Answer 1

Your ideas are nice, but some formulations should be improved.

1. "any sequence of functions $(f_n)_{n \in \Bbb N}$"
   $(f_n)_{n \in \Bbb N}$ is not a sequence of functions $f_n$, but a sequence of real numbers $f_n \in \mathbb R$. You can alternatively regard $(f_n)_{n \in \Bbb N}$ as a function $f : \mathbb N \to \mathbb R$.
   This misinterpretration occurs again in your second approach.

2. "so the sequence $(g_n)_{n \in \Bbb N}$ has only finitely many components that are bounded"
   The components are real numbers, thus the assertion does not make sense. What you mean is that the components $g_n$ of any $(g_n)_{n \in \Bbb N} \in B$ are bounded by some $M \ge 0$ (i.e. $\lvert g_n \rvert \le M$ for all $n$). But $(g_n)_{n \in \Bbb N} \in U_1 \times U_2 \times \dots \times \Bbb R \times \Bbb R \times \dots$ allows that $g_n \to \infty$ as $n \to \infty$.

3. "find a bounded sequence of functions $(f_n)_{n \in \Bbb N}$ that converges to non-bounded sequence of functions"
   Notation is confusing. In the previous part of your question you used $(f_n)_{n \in \Bbb N}$ to denote a single element of $\mathbb R^{\mathbb N}$, here you use it to denote it as a sequence of elements of $\mathbb R^{\mathbb N}$. I suggest to express it as follows: Find a sequence $(\phi_n)_{n \in \Bbb N}$ of elements of $\phi_n \in B$ (i.e. of bounded sequences of real numbers) which converges to an unbounded sequence $\phi \in \mathbb R^{\mathbb N}$. Your example is correct. The convergence in your example has of course to be proved. But it is a general theorem that a sequence $(x^n)$ of elements $x^n$ of a product $P = \prod_{i \in I} X_i$ of topological spaces $X_i$ converges to $x \in P$ iff all component-sequences $(x^n_i)$ (which are sequences in $X_i$) converge to the component $x_i$ of $x$. That is, convergence in a product means componentwise convergence.
   An alternative is to prove that $N = \mathbb R^{\mathbb N} \setminus B$ is not open in $\mathbb R^{\mathbb N}$. This can be done as in your first approach. Let $(f_n)_{n \in \Bbb N} \in N$. If $N$ were open, we would get a basic open $U = U_1 \times U_2 \times \dots \times U_k\times \Bbb R \times \Bbb R \times \dots$ such that $(f_n)_{n \in \Bbb N} \in U \subset N$. But we have $f = (f_1,\ldots,f_k,0,0,\ldots) \in U$, $f \notin N$, a contradiction.

Update:

You are indeed confusing a few things. In the first part you want to show that $B$ is not open in $X = \mathbb R^{\mathbb N}$. To do so, you have to pick an element $\xi$ of $B$ and to show that no open neigborhood $U$ of $\xi$ is contained in $B$. The elements of $X$ are functions $\mathbb N \to \mathbb R$, and you can denote them as you want, e.g. by $\xi : \mathbb N \to \mathbb R$ or by $f : \mathbb N \to \mathbb R$ or by something else. Let us use the $f$-notation. It is clear that $f$ can be identified with the real sequence $(f_n) = (f_n)_{n \in \Bbb N}$, where $f_n= f(n) \in \mathbb R$. It is just a notational issue which variant you prefer, but it seems to me that the sequence notation is more customary to denote elements of $X$. Now read again 1. above. The point $\xi \in B$ is not a sequence of functions $f_n : \mathbb N \to \mathbb R$, i.e. not a sequence of points of $X$, it is a sequence of real numbers.

By the way, picking a point of $B$ is unnecessary, it suffices to show that no basic open $U = U_1 \times U_2 \times \dots \times U_k\times \Bbb R \times \Bbb R \times \dots$ is contained in $B$. We have done this correctly by showing that $U$ contains unbounded sequences.

For your second approach you want to show that there exists a sequence in $B$ which converges to an element of $X \setminus B$. Sequences in $B$ are in fact sequences of functions $\mathbb N \to \mathbb R$, or sequences of real sequences. You can of course denote such a sequence by $(f_n)$, but in the light of 1. I find it confusing to use the same symbol $(f_n)$ both for elements of $X$ and for sequences in $X$.

As I said above, you can avoid to work with sequences in $X$ and show instead that $X \setminus B$ is not open. Again it suffices to show that no basic open $U = U_1 \times U_2 \times \dots \times U_k\times \Bbb R \times \Bbb R \times \dots$ is contained in $X \setminus B$. This is true because $U$ contains bounded sequences.

Update 2:

The shortest proof that $B$ is neither open nor closed in $X$ is this:

• Show that each basic open $U = U_1 \times U_2 \times \dots \times U_k\times \Bbb R \times \Bbb R \times \dots$ contains both bounded and unbounded real sequences (we have done this above).

• This means that each (basic) open set $U$ intersects both $B$ and $N = X \setminus B$, thus none of $B, N$ can be open.

Answer 2

Another approach would be to view the product topology as a metric space, as per Show that the countable product of metric spaces is metrizable

Then you have a space is closed if it contains all of its limit points. So to show it is not closed, we can construct a sequence of bounded sequences that converge to an unbounded sequence. This is easy, just let your sequence of sequences be the numbers 1 through $n$ in the first $n$ points and $0$ thereafter. Each sequence is clearly bounded, but it converges to the sequence $f_n=n$ which is unbounded.

A space is open if its complement is closed, so to show it is not open we can construct a sequence of unbounded sequences that converge to a bounded one. Here we can take as our sequence of sequences $f_n(i)=0$ if $n\not| i$ and $f(n)=n$ otherwise. Each is clearly unbounded, but it converges to the 0 sequence as the non-zero terms are pushed further and further down, which are each weighted at $2^{-n}$, and the 0 sequence is bounded